The document discusses uniform circular motion and centripetal acceleration. It defines uniform circular motion as motion in a circle with constant speed. For an object to travel in a circular path, it must be subjected to an unbalanced force and therefore experiences centripetal acceleration towards the center. Centripetal acceleration is calculated using the object's speed and the radius of the circular path. The centripetal force causing this acceleration can then be determined using Newton's Second Law.

1. Challenge Question
 Can an object be accelerated if its
speed remains constant?
 Yes - its direction can change.
 Can an object be accelerated if its
velocity remains constant?
• No… velocity is a vector quantity. If
velocity is constant, speed and direction
are constant.

2. Uniform Circular Motion
The Motion of an Object in a
Circle With a Constant or
Uniform Speed.

3. Uniform Circular Motion
 Since the direction of
the objects velocity is
changing, the object
must be subjected to
an UNBALANCED
FORCE!
 Therefore, it is
ACCELERATING

4. Uniform Circular Motion
 The direction of
the velocity
vector is the
same as the
direction of the
object's motion

5. Uniform Circular Motion
 The direction of the
acceleration is
inwards, always
pointing towards the
center of the circle
 Called “Centripetal
Acceleration”
 Centripetal means
“center-seaking”

6. Calculating Centripetal
Acceleration
 The centripetal acceleration (ac) of an object
in uniform circular motion is calculated by
means of the following equation:
ac = v2 / r (reference tables)
 v is the linear speed of the object
 r is the radius of the circular path

7. Calculating Centripetal Force
 The unbalanced force associated with
centripetal acceleration is called the
centripetal force (Fc). From Newton’s 2nd
Law we get:
Fc = m ac (reference tables)

8. Group Activity
How can we calculate the speed of an
object in UCM?
What do we need to measure?
What calculations do we need to make?

9. Calculating the Speed
 Can measure speed indirectly using a quantity
known as the period (T).
 The period (T) of revolution is the time an object
takes to complete one revolution
 In one revolution, the distance the object travels
equals the circumference of the circle (2r)

10. Calculating the Speed
 Using the equation
v = d / t, we substitute
and get…
 V = 2r / T

11. Practice Problem
 An object traveling in a circular path makes
1200 revolutions in 1.0 hour. If the radius
of the path is 10 meters, calculate the speed
of the object.

12. Solution
 If the object makes 1200 revolutions in 1.0
hr (3600 seconds) the time for one
revolution (Period T) is
 T = 3600 s / 1200 rev = 3 seconds
 V = 2r / T = (2) (3.14) (10 m) / (3.0 s)
 V = 21 m/s

13. Kepler’s Laws of Planetary
Motion
 Website Activity

14. Newton’s Law of Universal
Gravitation
 Newton recognized that the force
responsible for pulling an apple toward the
Earth has the same origin as the force that
keeps the Moon in its orbit around the
Earth.
 Force is called gravitation, and it is present
between all bodies of mass in the universe.

15. Newton’s Law of Universal
Gravitation
m1 m2
R
Fg
Fg
Fg = Gm1m2 / R2 (Reference Tables)

16. Newton’s Law of Universal
Gravitation
 The force of gravitation is an attractive
force.
 The law is known as an inverse square law.
 Constant G is called universal gravitational
constant. Its measured value is
6.67 x 10-11 N m2 / kg2 (reference tables)

17. Practice Problem
 Calculate the gravitational force between
the Earth and the Moon. Use your reference
tables to determine the mass of the Earth,
the mass of the Moon, and the distance
between the Earth and the Moon.

18. Are astronauts in orbit really
weightless?

19. Newton’s Law of Universal
Gravitation
 Video – “The Apple and the Moon” from
The Mechanical Universe.
 Worksheet – Satellite Motion and Weight
and Universal Gravitation

20. The Gravitational Field
A gravitational field is a region
of space that attracts masses
with a gravitational force.

21. The Gravitational Field
 The gravitational-field concept assumes that
one mass somehow changes the space
around it and a second mass then interacts
with the field.
 Copy schematic on the board

22. The Gravitational Field
 The gravitational field is a vector quantity
 Its direction is the direction of the force on
mass mo (called the test mass).
 The test mass is assumed to be much
smaller than the mass M.
 The gravitational field strength is the ratio
of the gravitational force on the test mass
(Fg) to the test mass (mo).

23. The Gravitational Field
 Therefore,
 g = Fg / mo
 Units of g are N / kg (same as m / s2)
 The gravitational field is simply another way of
viewing the action of gravity.
 Numerically, the gravitational acceleration and the
gravitational field strength are always equal.